Vectors: Examples

The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering. Each module will review main concepts, illustrate them with examples, and provide extensive practice problems.

審閱

TJ

Its a good way to start studying for the FE exam, but you will need to get a book with all the FE topics to study with as well.

PD

Dec 09, 2018

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These are best videos for the Working professional who can not devote much time reading the test material. Nicely explained !

從本節課中

Mathematics

This module reviews the basic principles of mathematics covered in the FE Exam. We first review the equations and characteristics of straight lines, then classify polynomial equations, define quadric surfaces and conics, and trigonometric identities and areas. In algebra we define complex numbers and logarithms, and show how to manipulate matrices and determinants. Basic properties of vectors with their manipulations and identities are presented. The discussion of series includes arithmetic and geometric progressions and Taylor and Maclaurin series. Calculus begins with definitions of derivatives and gives some standard forms and computation of critical points of curves, then presents grad, del and curl operators on scalar and vector functions. Differential equations are calcified and to methods to solve linear, homogenous equations are presented. Fourier series and transforms are defined along with standard forms, and finally Laplace transforms and their inverse are discussed. In all cases, basic ideas and equations are presented along with sample problems that illustrate the major ideas and provide practice on expected exam questions.Time: Approximately 4.5 hours | Difficulty Level: Medium

教學方

Dr. Philip Roberts

腳本

Now let's do some examples on vectors. So, the first example, we have a vector which is extending from the points with coordinates (1,1,3) to (5,4,1). So this is what it looks like, vector from (1,1,3) to (5,4,1) and couple of questions on this. First of all what is the magnitude of the vector between those two points? Which of these alternatives? So first we form the vector between those two points and vectors form very simply by taking the difference between the components of the second vector and the first one. So the x component is just 5- 1 the y component will be 4- 1 and the z component 1- 3. This equation here. So this equation and expanding that out we find that that is equal to 4i + 3j- 2k. And the magnitude of the vector is equal to the square root of the sum of squares of the components 4 squared + 3 squared + 2 squared = 5.39. So the answer is B. Next question, the unit vector is which of these? So, here what we want to find is a unit vector, in other words, a vector of length one extending between those two points. So, the general definition of the unit vector, m, is equal to the vector itself divided by the magnitude of the vector. And these two things we already know, here's the vector, 4i + 3j- 2k, and the magnitude of the vector, we've already calculated, is 5.39. So, computing out those coefficients, we find that that is equal to 0.74 + 0.56j- 0.37k, and the answer is A. And, you also might want to convince yourself, just add up the sum of the squares of those coefficients. Take the square root of it and convince yourself that the length or the magnitude of that vector is indeed 1. Next we have two vectors which are given by these two equations A is 4i + 2j + 3k and B = i + 5j. And the first question is, the sum of these two vectors is, which of these? So, the sum of the vectors, we can just add up all of the individual components here, so in this case, the i component is 4 + 1. The j component is 2+5 and the k component is 3+0. Which is the equation that which is given here. So expanding and computing that, we find that that is equal to 5i + 7j+ 3k and the answer is C. Next question, form the dot product between those two vectors. Which of these is it? And of course, this is a scalar quantity. So, we use our general relationship. The dot product A.B is equal to the sum of the products of the individual components. So, in this case, this is going to be 4 x 1, + 2 + 5, + 3 x 0. Which is this equation right here, that is equal to 14. So the answer is D. And lastly, we'll compute the cross product, in other words, the vector product, of these two vectors is which of these alternatives. So, this one is a little bit more complicated. And we do this by writing down the determinate i, j, k and then the two vectors a and b, and computing the determinant. So that is equal to the vector a is, has coefficients 4, 2, 3. And the vector b has coefficients 1, 5, and 0. So, to compute this, we start with the i term here, i, cross out the row and column. And that is equal to (2x0-3x5). Next, we form the j component. Cross that out and again because the signs alternate that's going to be -ga multiplied by 4 times 0- 3 times 1. And finally the k component which is going to be positive. 4 times 5- 2 times 1, expanding that out, we find that that is equal to -15i + 3j + 18k, and the answer is c. Last example, we have a parallelogram which is bounded by two vectors, which extend from the origin to the points 0 and 3, and 3 and 2. The area of the parallelogram that's formed by those two vectors is most nearly which of these alternatives? So, here is the situation. Here are the two vectors extending from the origin to 0 and 3, 0, 3 and 3, 2. And the question is, to find the area of the parallelogram. In other words, this area right here. And obviously there are a number of ways to do this, but in this case the easy way to do it is by forming the magnitude of the cross product between those two vectors. Because you remember that the cross product is just the the area of the parallelogram formed by the two vectors. So the first step is to write those two as vectors. And I'll call the vertical vector here A, and this vector, vector B. So the vector A is simply 0I plus 3J. The vector B is 3I plus 2J. So now I can form the cross product, same way again, ijk. And the vector A is 030. The vector B is 320. And when we compute this, maybe you can see right away that these vectors lie in the xy plane, so the only component that we are going to have here is the k component. In other words in the z direction, which is perpendicular to this plane. So, we know we are only going to get one component, but nevertheless, we'll compute them all just to illustrate how that works. So, the i component here, cross that out, is 3 times 0- 2 times 0, which is indeed, 0. The j component, is -0 times 0- 0 times 3, which is also 0, as we know. And finally, the k component, the only one which is going to be non-zero is equal to k times 0 times 3, 2- 3 times 3. In other words, it's equal to minus 9k, is the vector cross product. But here we're only concerned with the magnitude, so we don't care about the negative sign here, and the magnitude of that vector is equal to the area, which is equal to 9, so the answer is c. And this concludes our discussion of vectors.